1.

The decomposition of NO is studies by monitoring the concentration. Initially the concentration of NO is 2.33 moles/L and after 184 min it is reduced to 2.02 moles/L. The reaction takes place according to the equation 2NO to N_(2)+O_(2). Calculate the rate of production of N_(2).

Answer»

Solution :`2NO implies N_(2)+O_(2)`
Rate of DECOMPOSITION of NO `=(DELTA[NO])/(Delta t)=([2.33-2.02])/(184xx60)`
`=2.8xx10^(-5) "mole"//L//s`
Rate of PRODUCTION of `N_(2)=(1)/(2)xx ` rate of decomposition of NO
`therefore` Rate of production of `N_(2)=(2.8xx10^(-5))/(2)=1.4xx10^(-5) " mole" //L//s`.


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